andria wrote: ... let the line parallel to $BC$ intersect circumcircle of $\triangle ABC$ at $A_1$.... Paralell from where??

I'm sorry the parallel is from $ A $ I edit it.

Seems hard... Nice problem!!

My solution: Lemma: Let $\triangle ABC$ be a triangle with incenter $I$. the line parallel with $BC$ trough $A$ intersects the circumcircle of $\triangle ABC$ at $X$. Let $M$ be the intersection of $AI$ with the circumcircle. $R$ and $S$ are the intersections of lines $XI$ and $XM$ with $BC$ respectively. Then we have $IR$=$IS$. Proof: $D$ is the intersection of $AM$ with $BC$. $N$ is the intersection of the line parallel with $BC$ trough $I$ and $XM$. easy to see that $IS$=$DN$. So we have to prove that $INDR$ is a parallelorgam or $IN=RD$. Let $Q$,$O$ and $P$ be the intersections of the perpendicular lines trough $I$ to $BC$,$AX$ and $AC$ with them respectively.$IPOA$ is cyclic. Then we heve $\frac{IN}{AX}=\frac{IM}{AM}=\frac{MC}{MX}=\frac{sin(\angle IAP)}{sin(\angle IAO)}=\frac{IP}{IO}=\frac{IQ}{IO}=\frac{RD}{AX}$. So $IN=RD$. Back to the main problem: Let $M$ be the intersection of $AI$ with the circumcircle. Using the Lemma we get that the lines $A_{1}E$,$A_{1}I$,$A_{1}M$ and $A_{1}A$ are harmonic. With angle chasing we get that $A_{1}M$ is tangent to the circumcircle of $\triangle AA_{1}T$. So $A_{1}I$ is the symmedian of $\angle AA_{1}T$. And from this $\angle TAI=\angle TED=\angle AA_{1}T$ we get that $I$ is the intersection of the tangent lines to the circumcircle of $\triangle AA_{1}T$ at $T$ and $A$. So $AI=TI$. Source:Iran TST 2015,second exam,first day,problem 2

My solution: $M, S$ are the midpoints of $BC$ and $\overarc{BC}$ not containing $A$ $F$ is the tangent point of $BC$ and the A-excircle $\Rightarrow F = S_M(E)$ The reflection WRT the perpendicular bisector of $BC$ maps: $A\mapsto{A_1}, F\mapsto{E}$ $\Rightarrow G = AF\cap{A_1E} \in (OMS)$, which is the perpendicular bisector of $BC$ On the other hand: it’s well-known that $IM\parallel{AF} (\because AF$ passes through Nagel point$)$ $\Rightarrow IM\parallel{AG}$ $\Rightarrow \frac{SM}{SG} = \frac{SI}{SA} = \frac{SD}{SI}$ $\Rightarrow MD\parallel{GI}$ $\Rightarrow \angle{TAI} \equiv \angle{TAD} = \angle{TED} = \angle{IGE}$ $\Rightarrow ATGI$ is cycic (1) On the other hand: $\triangle{GAA_1}$ is isoceles, $GI\parallel{AA_1}$ $\Rightarrow GI$ is an external bisector of $\angle{AGT}$ (2) (1), (2) $\Rightarrow IA = IT$ Q.E.D

My solution : Let $ X \equiv A_1E \cap \odot (ABC) $ . Since $ X $ lie on the isogonal conjugate of A-Nagel line of $ \triangle ABC $ WRT $ \angle A $ , so $ X $ is the tangency point of $ \odot (ABC) $ with A-mixtilinear incircle of $ \triangle ABC $ , hence $ XI $ is the bisector of $ \angle CXB $ (well-known) $ \Longrightarrow XI $ is the bisector of $ \angle A_1XA \equiv \angle TXA $ . ... $ (\star) $ Since $ \angle ATX=\angle ADE=90^{\circ}-\tfrac{1}{2} | \angle CBA-\angle ACB |=90^{\circ}-\tfrac{1}{2} \angle TXA $ , so $ \triangle XAT $ is an isosceles triangle with base $ AT $ $ \Longrightarrow XI $ is the perpendicular bisector of $ AT $ (from $ (\star ) $) $ \Longrightarrow AI=TI $ . Q.E.D

Let $A_1E $ intersect $(ABC) $ at $ L$ it s easy to show by angle chasing that $EDLS$ is cyclic and $ LI$ pass through $S'$ the diametrical opposite of the point $S$. applying REIM theorem to $ (LSDE)$ and $DEAT$ yields $ AT \parallel LS $ then $LI \perp AT$ but $LI=LS'$ is bisector of $\widehat{ALA_1}=\widehat{ALT} $ thus it s the bisector of $AT$

PROF65 wrote: Let $A_1E $ intersect $(ABC) $ at $ L$ it s easy to show by angle chasing that $EDLS$ is cyclic and $ LI$ pass through $S'$ the diametrical opposite of the point $S$. applying REIM theorem to $ (LSDE)$ and $DEAT$ yields $ AT \parallel LS $ then $LI \perp AT$ but $LI=LS'$ is bisector of $\widehat{ALA_1}=\widehat{ALT} $ thus it s the bisector of $AT$ Why $ LI$ pass through $S'$ the diametrical opposite of the point $S$ ?

i think i know where this problem come from : $lemma1:$ let $M$ be the mixtilinear point of $A$ and $D$ the foot of $I$ to $BC$ and $A_1$ the reflection of $A$ about perpendicular bisector of $BC$ . then $D$ $M$ $A_1$ are collinear $lemma2:$ $MI$ passes through $T'$ which both can be proved by considering the fact that $\widehat{BAM}=\widehat{CAX_a}$

5 months ago I didn't want to try this problem after seeing the above solution. . Iran TST 2015 P8 wrote: In triangle $ABC$(with incenter $I$) let the line parallel to $BC$ from $A$ intersect circumcircle of $\triangle ABC$ at $A_1$ let $AI\cap BC=D$ and $E$ is tangency point of incircle with $BC$ let $ EA_1\cap \odot (\triangle ADE)=T$ prove that $AI=TI$. It's well known that $A_1E\cap\odot(ABC)=T_A$ is the $A-\text{Mixtillinear Incircle}$ touch point. For the Prrof Refer to Fact 1 over. here. Let $T_AI\cap \odot(ABC)=X$. So, $X$ is the midpoint of $\widehat{BAC}$. So, $T_AI$ bisects $\angle AT_AT$. Also it's well known that $\angle AT_AB=ET_AC=\angle C\implies \angle AT_AT=180^\circ-\angle A-2\angle C$ and $\angle ATE=\angle ADB=\angle C+\frac{\angle A}{2}\implies\angle T_AAT=\angle C+\frac{\angle A}{2}\implies XT_A$ is the perpendicular bisector of $AT$. So, $AI=TI$. $\blacksquare$

Let $F$ be the tangential point of the $A-excircle$ with $BC$. It is well-known that $E$ and $F$ are symmetrical w.r.t. the midpoint of $BC$. Therefore $AA_1FE$ is an isoceles trapezoid. As a result, applying sprial sim. lemma to circle $(AEDT)$ and $(AEFA_1)$ we have $$\triangle ATA_1\sim\triangle ADF$$Therefore $\angle IAT=\angle DAT=\angle FAA_1=\angle AFD$ Let $\angle AFD=\alpha$. To show that $IA=IT$ it suffices to show that \begin{align} \frac{AT}{2AI}&=\cos\alpha \end{align}Now from the similarity we see that the $L.H.S.$ of $(1)$ is $$\frac{AA_1\cdot AD}{2AF\cdot AI}$$. Now notice that $$\frac{AA_1}{AB}=\frac{\sin(B-C)}{\sin C}$$$$\frac{AD}{AI}=\frac{AD}{AB}\frac{AB}{AI}=\frac{\sin B\cos\frac{C}{2}}{\sin(B+\frac{A}{2})\sin\frac{B}{2}}$$$$\frac{AF}{AB}\cos\alpha=\frac{\sin B}{\sin\alpha}\cos\alpha=\sin B\cot\alpha$$Combining all these our equation becomes \begin{align*} \frac{\sin(B-C)\sin B\cos\frac{C}{2}}{2\sin C\sin(B+\frac{A}{2})\sin\frac{B}{2}}=\sin B\cot\alpha \end{align*}cancelling terms using double angle forumla, it suffices to prove $$\frac{\sin(B-C)}{4\sin \frac{C}{2}\sin(B+\frac{A}{2})\sin\frac{B}{2}}=\cot\alpha$$Now let $EI$ meet $AF$ at $G$, it is well-known that $EG=2r$ where $r$ is the inradius. Therefore $$\cot\alpha=\frac{EF}{2r}=\frac{a-2(s-c)}{2r}=\frac{c-b}{2r}=\frac{(s-b)-(s-c)}{2r}=\frac{1}{2}(\cot\frac{C}{2}-\cot\frac{B}{2})$$Hence it suffices to prove $$\frac{\sin(B-C)}{2\sin(B+\frac{A}{2})}=\sin\frac{C}{2}\sin\frac{B}{2}(\cot\frac{C}{2}-\cot\frac{B}{2})$$By compound angle formula the right hand side is $\sin(\frac{B-C}{2})$. Hence it suffices to prove $$\cos{\frac{B-C}{2}}=\sin(B+\frac{A}{2})$$which is trivial since $90^{\circ}+\frac{B-C}{2}=B+\frac{A}{2}$

Let $N$ be the midpoint of arc $BAC$, let $M_{BC}$ be the midpoint of arc $BC$, let $\omega$ be the circle with center $N$ and radius $NA$, note that $A'\in \omega$. Let $T'=\omega\cap A'E$. Let $P=(ABC)\cap A'E$. We also use the following facts that $A',E,P$ are collinear and $P,I,N$ are collinear. We claim that $T\equiv T'$. $$\measuredangle AT'E=\measuredangle AT'A'=\frac{\measuredangle ANA'}{2}=\measuredangle ANM_{BC}= \measuredangle ACB+\measuredangle BAD=\measuredangle ADE,$$therefore $T\equiv T'$. Also, $\measuredangle PAT=\measuredangle PTA+\measuredangle APT=\measuredangle ANA'-\measuredangle ATE=\measuredangle ATP$, hence we conclude that $NP$ is the perpendicular bisector of $AT$, hence $IA=IT$.

[asy][asy]import olympiad;import geometry; size(9cm); defaultpen(fontsize(10pt));pen sec=mediummagenta;pen cho=pink;pen ges=deepmagenta; pair O,A,B,C,I,D,E,M,a,T,P,N; O=(0,0);A=dir(130);B=dir(200);C=dir(340);path abc=circumcircle(A,B,C); I=incenter(A,B,C); E=foot(I,B,C);D=extension(A,I,B,C);M=midpoint(B--C);a=2*foot(A,O,M)-A;path w=circumcircle(A,E,D);T=intersectionpoints(w,a--E)[0];P=intersectionpoints(abc,3*E-2a--a)[1];N=intersectionpoints(abc,4*I-3*P--P)[0]; draw(abc);draw(w);draw(A--B--C--cycle);draw(incircle(A,B,C));draw(I--E);draw(A--D);draw(P--a);draw(N--P);draw(A--T);draw(circle(N,abs(N-A))); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$E$",E,dir(E)); dot("$D$",D,dir(D)); dot("$A'$",a,dir(a)); dot("$T$",T,dir(T)); dot("$P$",P,dir(P)); dot("$N$",N,dir(N)); [/asy][/asy]

Idk what I’m doing now Let $T_a$ be the $A$-mixtilinear point of $\triangle ABC$, $M_a$ the arc midpoint of $BC$ not containing $A$ on $(ABC)$, $N_a$ the arc midpoint of $BAC$ on $(ABC)$, and $T’$ the reflection of $A$ over $\overline{N_aIT_a}$ (well-known fact that $N_a-I-T_a$). I claim now that $T=T’$ which suffices; to do this I’ll prove two things: $T_a-E-T’-A_1$ (well-known fact that $T_a-E-A_1$) $T’\in(ADE)$ which suffices. We first prove the first one. $\measuredangle N_aT_aT=\measuredangle AT_aN_a=\measuredangle AA_1N_a=\measuredangle N_aAA_1=\measuredangle N_aT_aA_1$, which wins. The second one follows from the fact that $\overline{AT}\parallel\overline{T_aM_a}$ as they are both perpendicular to $\overline{N_aIT_a}$, so $\measuredangle AT’E=\measuredangle M_aT_aE=\measuredangle M_aDE=\measuredangle ADE$ as desired.